Question

A bi-convex lens is formed with two thin plano convex lenes . Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R = 14 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be?

-280.0 cm

40.0 cm

21.5 cm

13.3 cm

Answer 1

P = (1.5-1) (1/14-0) + (1.2-1) (0-1/(-14))

P = 1/20

f=+20cm

1/v-1/(-40) = 1/20

v = 40cm

Hope this helps...   : )

Answer 2

Given refractive index of lens 1 n₁ = 1.5 and  

refractive index of lens 1 n₂ = 1.2

Radius of curvature R₁ = 14 cm

Object distance u = - 40 cm

We know,

(1/f) = (μ -1) ( 1/R₁ - 1/R₂)

Thus the combined focal length will be

(1/f) = (1.5 -1)(1/14 - 1/∞) +(1.2 - 1)( 1/14 - 1/∞)

(1/f) = (0.5/14)+(0.2/14)

f =20 cm

From lens formula we have,

(1/v) - (1/u) =(1/f)

(1/v) = (1/f) + (1/u)

(1/v) = (1/20) -(1/40)

v = 40 cm